Simplify and expand the following expression: $ \dfrac{1}{4x + 24}+ \dfrac{3}{3x - 27}- \dfrac{1}{x^2 - 3x - 54} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{1}{4x + 24} = \dfrac{1}{4(x + 6)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3x - 27} = \dfrac{3}{3(x - 9)}$ We can factor the quadratic in the third term: $ \dfrac{1}{x^2 - 3x - 54} = \dfrac{1}{(x + 6)(x - 9)}$ Now we have: $ \dfrac{1}{4(x + 6)}+ \dfrac{3}{3(x - 9)}- \dfrac{1}{(x + 6)(x - 9)} $ The least common multiple of the denominators is: $ 12(x + 6)(x - 9)$ In order to get the first term over $12(x + 6)(x - 9)$ , multiply by $\dfrac{3(x - 9)}{3(x - 9)}$ $ \dfrac{1}{4(x + 6)} \times \dfrac{3(x - 9)}{3(x - 9)} = \dfrac{3(x - 9)}{12(x + 6)(x - 9)} $ In order to get the second term over $12(x + 6)(x - 9)$ , multiply by $\dfrac{4(x + 6)}{4(x + 6)}$ $ \dfrac{3}{3(x - 9)} \times \dfrac{4(x + 6)}{4(x + 6)} = \dfrac{12(x + 6)}{12(x + 6)(x - 9)} $ In order to get the third term over $12(x + 6)(x - 9)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{1}{(x + 6)(x - 9)} \times \dfrac{12}{12} = \dfrac{12}{12(x + 6)(x - 9)} $ Now we have: $ \dfrac{3(x - 9)}{12(x + 6)(x - 9)} + \dfrac{12(x + 6)}{12(x + 6)(x - 9)} - \dfrac{12}{12(x + 6)(x - 9)} $ $ = \dfrac{ 3(x - 9) + 12(x + 6) - 12} {12(x + 6)(x - 9)} $ Expand: $ = \dfrac{3x - 27 + 12x + 72 - 12}{12x^2 - 36x - 648} $ $ = \dfrac{15x + 33}{12x^2 - 36x - 648}$ Simplify: $ = \dfrac{5x + 11}{4x^2 - 12x - 216}$